The discussion was predicated on this reaction as follows: Si28 + O16 -> Ca40 + He4 + 6.2 MeV
Now you are changing the playing field to U236 fission. Let us continue to talk about the fusion of highly stable elements. On Sat, Jul 13, 2013 at 3:35 AM, <[email protected]> wrote: > In reply to Axil Axil's message of Fri, 12 Jul 2013 19:43:21 -0400: > Hi, > [snip] > >What causes the defeat of coulomb barrier? > > > > > > > > > > > >Mixent - kinetic energy of nuclear collision. > > > > > > > > > > > >Axil – EMF- namely charge concentration and anapole magnetism. > > > > > > > > > > > > > > > >The defeat of the coulomb barrier can’t be kinetic energy because from the > >experience gained with the kinetic collisions of nuclei we know that the > >resultant nucleus would be left excited by the kinetic energy of that > >nuclear collision and affect the combined nucleus by exciting it leaving a > >radioactive isotope as an end product. > > > > > > > > > > > >Internal rearrangement of the combined nuclei by EMF does not carry excess > >energy of collision into the reaction therefore no radioactive isotope is > >produced > > Kinetic energy only contributes to the overall energy of the reaction. What > comes out is not necessarily related to the fact that kinetic energy > played a > role. E.g. conventional fission in U235 is best triggered by a slow moving > neutron with as little kinetic energy as possible, since low kinetic energy > enhances the cross section. > Despite the lack of initial kinetic energy, the result is usually a > plethora of > radioactive nuclei. The primary reason for this is simply that relative to > its > lighter daughter nuclei, U235 has an excess of neutrons. That means that > the > daughter nuclei are neutron rich, and hence radioactive. > > When light nuclei combine at high speed, the resultant momentary "heavy" > nucleus > is neutron poor (compared to normal stable nuclei of the same combined > weight), > hence the immediate fission nuclei from the combined nucleus will not be > neutron > rich, and hence not likely to be radioactive. > > If it doesn't have to, nature actually prefers not to make radioactive > nuclei. > > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > >
